Question

For certain workers, the mean wage is $5.00/hr, with a standard deviation of $0.25. If a worker is chosen at random, what is the probability that the workers wage is between $4.25-$5.75. Assume a normal distribution of wages.

Answer 1

The probability is 0.9973 or 99.73%

Explanation:

* Lets explain how to solve the problem

- For the probability that a < X < b (X is between two numbers, a and b),

convert a and b into z-scores and use the table to find the area

between the two z-values.

- Lets revise how to find the z-score

- The rule the z-score is z = (x - μ)/σ , where

# x is the score

# μ is the mean

# σ is the standard deviation

* Lets solve the problem

- For certain workers, the mean wage is $5.00/hr, with a standard

deviation of $0.25

∴ μ = 5 and σ = 0.25

- The worker wage is between $4.25 and $5.75

∴ 4.25 < X < 5.75

∵ z = (x - μ)/σ

∴ z = (4.25 - 5)/0.25 = -0.75/0.25 = -3

∴ z = (5.75 - 5)/0.25 = 0.75/0.25 = 3

- Use the z table to find the corresponding area

∵ P(z > -3) = 0.00135

∵ P(z < 3) = 0.99865

∴ P(-3 < z < -2) = 0.99865 - 0.00135 = 0.9973

∵ P(4.25 < X < 5.75) = P(-3 < z < 3)

∴ P(4.25 < X < 5.75) = 0.9973 = 99.7%

* The probability is 0.9973 or 99.73%

Answer 2

P(-3.0 < z < 3.0) = 0.9974

Explanation:

Mean = 5

Standard Deviation = 0.25

We need to find P(4.25 <x<5.75)

z = x - mean/standard deviation

z = 4.25 - 5/0.25

z = -3.0

z = x - mean/standard deviation

z = 5.75 - 5/0.25

z = 3.0

So, P(4.25 <x<5.75) P(-3.0 < z < 3.0)

Finding values from the z-score table

P(z<-3.0) = 0.0013

P(z<3.0) = 0.9987

P(-3.0 < z < 3.0)=P(z<3.0) - P(z<-3.0)

P(-3.0 < z < 3.0) = 0.9987 - 0.0013

P(-3.0 < z < 3.0) = 0.9974